We are asked to find the area of the region bounded by `f(x)=-x^2+4x` and `y=0` .
We can factor the quadratic to y=-x(x-4); the x-intercepts are 0 and 4.
The equivalent question then is to find the definite integral `int_0^4 (-x^2+4x)dx` .
`int_0^4 (-x^2+4x)dx=-x^3/3+2x^2 |_0^4`
`=(-64/3+32)-(0+0)=32/3 `
So the area bounded by the given curve and the x-axis is 32/3.
An alternative approach is to use a Riemann sum:
The area can be found by `lim_{n->oo} sum_{i=1}^{n} f(c_{i})Delta x_{i}`
If we use a regular partition and choose the right-hand endpoint in each subinterval we get:
`A=lim_{n->oo} sum_{i=1}^{n} (-((4i)/n)^2+4((4i)/n))(4/n) `
`=lim_{n->oo}[(-64)/(n^2)sum_{i=1}^n(i^2/n-i)] `
`=lim_{n->oo}[(-64)/n^2(1/n((n(n+1)(2n+1))/6)-(n(n+1))/2)] `
`=lim_{n->oo}[(-64)/n^2(1/3n^2+1/2n+1/3-n^2/2-n/2)] `
`=lim_{n->oo}[(-64)/3-(32)/n-(64)/n^3+32+(32)/n] `
`=(-64)/3+32=32/3=10.bar(6) `
The graph: